Difference between revisions of "2017 AMC 12B Problems/Problem 15"
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Therefore, our answer is <math>\boxed{\textbf{(E) }37:1}</math>. | Therefore, our answer is <math>\boxed{\textbf{(E) }37:1}</math>. | ||
− | ==Solution 2: Inspection== | + | ==Solution 2: Inspection(easiest solution)== |
Note that the height and base of <math>\triangle A'CC'</math> are respectively 4 times and 3 times that of <math>\triangle ABC</math>. Therefore the area of <math>\triangle A'CC'</math> is 12 times that of <math>\triangle ABC</math>. | Note that the height and base of <math>\triangle A'CC'</math> are respectively 4 times and 3 times that of <math>\triangle ABC</math>. Therefore the area of <math>\triangle A'CC'</math> is 12 times that of <math>\triangle ABC</math>. | ||
Revision as of 23:07, 30 December 2019
Contents
Problem 15
Let be an equilateral triangle. Extend side beyond to a point so that . Similarly, extend side beyond to a point so that , and extend side beyond to a point so that . What is the ratio of the area of to the area of ?
Solution 1: Law of Cosines
Solution by HydroQuantum
Let .
Recall The Law of Cosines. Letting , Since both and are both equilateral triangles, they must be similar due to similarity. This means that .
Therefore, our answer is .
Solution 2: Inspection(easiest solution)
Note that the height and base of are respectively 4 times and 3 times that of . Therefore the area of is 12 times that of .
By symmetry, . Adding the areas of these three triangles and for the total area of gives a ratio of , or .
Solution 3: Coordinates
First we note that due to symmetry. WLOG, let and Therefore, . Using the condition that , we get and . It is easy to check that . Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is
Solution by mathwiz0803
Solution 4: Computing the Areas
Note that angle is °, as it is supplementary to the equilateral triangle. Then, using area and letting side for ease, we get: as the area of . Then, the area of is , so the ratio is
Solution by Aadileo
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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